A set of positive integers satisfies the property that when 10^{20} , 15^{10} and 24^{15} are divided by any number in this set, at least one of the remainders is zero. What is the total number of elements in this set? a)1256 b)1266 c)1024 d)none of these

## Comments

Sort by:

TopNewestimage

Let

Adenote numbers divisible by \(10^{20}=2^{20}5^{20}\),Bdenote numbers divisible by \(15^{10}=3^{10}5^{10}\) andCdenote numbers divisible by \(24^{15}=2^{45}3^{15}\).What you want is \(a+b+c+d+e+f+g\).

g=Numbers divisible by all 3. So g has one number {1} so g=\(\color{blue}{1}\)

d+g=Numbers in A and C= divisible by \(2^{20}\). Hence d+g=21 or d=\(\color{blue}{20}\)

f+g=Numbers in B and C= divisible by \(3^{10}\). Hence f+g=11 or f=\(\color{blue}{10}\)

e+g=Numbers in A and B= divisible by \(5^{10}\). Hence e+g=11 or e=\(\color{blue}{10}\)

a+d+e+g=Numbers in A=divisible by \(10^{20}\). a+d+e+g=441 or a=\(\color{blue}{410}\)

b+f+e+g=Numbers in B=divisible by \(15^{10}\). b+f+e+g=121 or b=\(\color{blue}{100}\)

c+d+f+g=Numbers in C=divisible by \(24^{15}\). c+d+f+g=736 or c=\(\color{blue}{705}\)

Adding, we get a+b+c+d+e+f+g=\(\color{red}{\boxed{1256}}\) – Pranjal Jain · 2 years, 4 months ago

Log in to reply